Two forces 10N and 20N are inclined at an angle of 60 to each other.find the resultant force?if the two forces are now made to be inclined at 120 to each other.find the magnitude of the resultant forces.
120*10
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To find the resultant force when two forces are inclined at an angle of 60 degrees to each other, we can use the law of cosines. The formula is:
resultant force = sqrt(F1^2 + F2^2 + 2(F1)(F2)cosθ)
where F1 and F2 are the magnitudes of the two forces, and θ is the angle between them.
Using this formula, let's calculate the resultant force:
F1 = 10N
F2 = 20N
θ = 60 degrees
resultant force = sqrt((10)^2 + (20)^2 + 2(10)(20)cos(60))
Using the formula for cos(60) = 0.5, we can simplify the equation:
resultant force = sqrt(100 + 400 + 400) = sqrt(900) = 30N
Therefore, the magnitude of the resultant force when the two forces are inclined at 60 degrees to each other is 30N.
Now, let's calculate the magnitude of the resultant force when the two forces are inclined at 120 degrees to each other.
F1 = 10N
F2 = 20N
θ = 120 degrees
resultant force = sqrt((10)^2 + (20)^2 + 2(10)(20)cos(120))
Using the formula for cos(120) = -0.5, we can simplify the equation:
resultant force = sqrt(100 + 400 - 400) = sqrt(100) = 10N
Therefore, the magnitude of the resultant force when the two forces are inclined at 120 degrees to each other is 10N.
To find the resultant force, we can use the concept of vector addition. The resultant force is the vector sum of the two forces.
1. When the two forces are inclined at an angle of 60 degrees to each other:
We can break down each force into its components along the x-axis and y-axis, using trigonometry. Let's call F1 the magnitude of the first force (10N) and F2 the magnitude of the second force (20N).
The x-component of F1, denoted as F1x, can be calculated as:
F1x = F1 * cos(60) = 10N * cos(60) = 10N * 0.5 = 5N
The y-component of F1, denoted as F1y, can be calculated as:
F1y = F1 * sin(60) = 10N * sin(60) = 10N * 0.866 = 8.66N
Similarly, for the second force:
The x-component of F2, denoted as F2x, is:
F2x = F2 * cos(60) = 20N * cos(60) = 20N * 0.5 = 10N
The y-component of F2, denoted as F2y, is:
F2y = F2 * sin(60) = 20N * sin(60) = 20N * 0.866 = 17.32N
Now, to find the resultant force, we add the corresponding components together:
Rx = F1x + F2x = 5N + 10N = 15N
Ry = F1y + F2y = 8.66N + 17.32N = 25.98N
Using the Pythagorean theorem, we can find the magnitude of the resultant force:
R = √(Rx^2 + Ry^2) = √(15N^2 + 25.98N^2) ≈ √(225N^2 + 674.40N^2) ≈ √(899.40N^2) ≈ 29.98N
Therefore, when the two forces are inclined at an angle of 60 degrees to each other, the magnitude of the resultant force is approximately 29.98N.
2. When the two forces are inclined at an angle of 120 degrees to each other:
We can follow the same procedure as above to find the components of the forces along the x-axis and y-axis. However, since the angle is now 120 degrees, we need to adjust the signs in the calculations.
F1x = F1 * cos(120) = -5N
F1y = F1 * sin(120) = 8.66N
F2x = F2 * cos(120) = -10N
F2y = F2 * sin(120) = -17.32N
Now, adding the x-components and y-components together:
Rx = F1x + F2x = -5N + (-10N) = -15N
Ry = F1y + F2y = 8.66N + (-17.32N) = - 8.66N
Again, using the Pythagorean theorem:
R = √(Rx^2 + Ry^2) = √((-15N)^2 + (-8.66N)^2) = √(225N^2 + 75.03N^2) = √(300.03N^2) ≈ 17.32N
Therefore, when the two forces are inclined at an angle of 120 degrees to each other, the magnitude of the resultant force is approximately 17.32N.